I want to solve an equation for wavelength, but my algebra is a little rusty. Given the following equation:$$\tilde{\nu}=\frac{1}{\lambda}=R\bigg(\frac{Z^2}{n^2_i}-\frac{Z^2}{n^2_i}\bigg)$$
Is there a better way to solve for λ than this?
$$\lambda=\frac{1}{\nu}=\frac{1}{R(Z^2/n^2_i-Z^2/n^2_f)}$$

Question

I want to solve an equation for wavelength, but my algebra is a little rusty.  Given the following equation:$$\tilde{\nu}=\frac{1}{\lambda}=R\bigg(\frac{Z^2}{n^2_i}-\frac{Z^2}{n^2_i}\bigg)$$
Is there a better way to solve for λ than this?
$$\lambda=\frac{1}{\nu}=\frac{1}{R(Z^2/n^2_i-Z^2/n^2_f)}$$

anonymous · Answer

It seems like I should be able to have everything on the either the top or bottom of the fraction, instead of embedding fractions.

anonymous · Answer

I plugged this problem into WolframAlpha, and it gave me an idea for simplifying it.  Does this work?$$\lambda = \frac{1}{\tilde \nu}=-\frac{n_i^2n_f^2}{RZ^2(n_i^2-n_f^2)} = -\bigg(\frac{1}{R(n_i^2-n_f^2)}\bigg)\bigg(\frac{n_in_f}{Z}\bigg)^2$$

anonymous · Answer

The above is correct, however, I talked to the people in the math OpenStudy group, and evidently this is the most compact way to write the formula to solve for the wavelength in the Bohr model:
$$λ=\tilde \nu^{−1}=(RZ^2(n^{−2}_i−n^{−2}_f))^{−1}$$